Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $t = \dfrac{4(3y + 1)}{-8} \times \dfrac{4y}{21y^2 + 7y} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $t = \dfrac{ 4(3y + 1) \times 4y } { -8 \times (21y^2 + 7y) } $ $ t = \dfrac {4y \times 4(3y + 1)} {-8 \times 7y(3y + 1)} $ $ t = \dfrac{16y(3y + 1)}{-56y(3y + 1)} $ We can cancel the $3y + 1$ so long as $3y + 1 \neq 0$ Therefore $y \neq -\dfrac{1}{3}$ $t = \dfrac{16y \cancel{(3y + 1})}{-56y \cancel{(3y + 1)}} = -\dfrac{16y}{56y} = -\dfrac{2}{7} $